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Alborosie
1 year ago
9

What is the range of possible sizes for side x?

Mathematics
1 answer:
posledela1 year ago
4 0

Answer:

x = 3.96

Step-by-step explanation:

apply the pitagoras theorem

x =

\sqrt{ {2.8}^{2}  +  {2.8 }^{2} }

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999^2-998^2<br> Explain w/ steps.<br><br> Thank you
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Answer:

1997

Step-by-step explanation:

999^2 = 999 x 999 = 998,001

998 ^ 2 = 998 x 998 = 996,004

998,001 - 996, 004 = 1997

Hope this helps

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A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean l
andrew11 [14]

Answer:

86.65% probability that the mean monitor life would be greater than 74.4 months in a sample of 85 monitors

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

Population: \mu = 75, \sigma = 5

Sample of 85: n = 85, s = \frac{5}{\sqrt{85}} = 0.5423

If the claim is true, what is the probability that the mean monitor life would be greater than 74.4 months in a sample of 85 monitors

This is 1 subtracted by the pvalue of Z when X = 74.4. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{74.4 - 75}{0.5423}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

86.65% probability that the mean monitor life would be greater than 74.4 months in a sample of 85 monitors

6 0
2 years ago
What is 120.571 in expanded form?
gladu [14]
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